Let `a_1, a_2, a_3, ,a_n` be in G.P. such that `3a_1+7a_2+3a_3-4a_5=0.` Then common ratio of G.P. can be `2` b. `3/2` c. `5/2` d. `-1/2`

Let a_1,a_2,a_3 ......, a_n are in A.P. such that a_n = 100 , a_40-a_39=3/5 then 15^(th) term of A.P. from end is

Let a_1,a_2," ..."a_10 be a G.P. If a_3/a_1=25," then "a_9/a_5 equals

If a_1,a_2,a_3,. . . ,a_n. . . are in A.P. such that a_4−a_7+a_10=m , then sum of the first 13 terms of the A.P.is

If the first term of a G.P.a_(1),a_(2),a_(3),.. is unity such that 4a_(2)+5a_(3) is least then the common ratio of G.P.is -2/5 b.-3/ c./5 d.none of these

"If "a_1,a_2,a_3,.....,a_n" are in AP, prove that "a_(1)+a_(n)=a_(r)+a_(n-r+1)""

Let a_1,a_2 ,a_3 …..be in A.P. and a_p, a_q , a_r be in G.P. Then a_q : a_p is equal to :

Let a_1,a_2,a_3,... be in A.P. With a_6=2. Then the common difference of the A.P. Which maximises the product a_1a_4a_5 is :

Evaluate: /_\ |[1+a_1, a_2, a_3],[a_1, 1+a_2, a_3],[a_1, a_2, 1+a_3]|